Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: SkiPY on October 25, 2007, 06:18:34 AM
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Hi, I have a problem, I want to calculate a concentration of ethylalcohol (C2H5OH) in air. But i have ethylalcohol in liquid phase. So i need to evaporate ethylalcohol and after calculate it's concentration in air (maybe from partial pressures of both - air and ethylalcohol). But how?
temperature 22°C
partial pressure of ethylalcohol at 22°C = 6,6 kPa
atmospheric pressure of air = 101,325 kPa
capacity of air = 5L = 5 dm3
amount of ethylalcohol in liquid phase = ? ml
desired concentration: 2000ppm
Thank's in advance.
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You need also a density of ethanol. Not all ethanol will evaporate.
Note 1 ppm is equal to 1 mg of ethanol in 1 kg of air.
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AWK thanks for reply, but I know what is part per milion (ppm). I've found some useful sites which can help me with calculation, but i don't know if my method is alright.
http://www.lenntech.com/calculators/converter-parts-per-million.htm
http://homedistiller.org/calc.htm
I've found out, that 1 mL of ethanol = 0.789 g
From the mentioned ppm calculator: Total Molecular Weight of ethanol is 46.07g/mol
So from my computation: to get 2000ppm (in 5L of air) i need 0,020285 mg of ethanol = 0,0257 ul (microliter) of ethanol.
What do you think ?
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Note, not all ethanol evaporate. You should use a partial pressure for checking part of ethanol that evaporated.
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Note, not all ethanol evaporate. You should use a partial pressure for checking part of ethanol that evaporated.
but i've changed the input requirements (see 1st post). So if I calculating the amount of liquid to get 2000ppm, this amoun should be surely evaporated. ???
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1 ppm = 1 mg of ethanol /1kg of air. Mass of 5 l of air is about 6.45 g = 0.00645 kg
hence 2000 ppm is equal to 12.9 mg of ethanol
otherwise you should state you use noncoventional ppm units, eg mg/liters_of_air
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I would calculate ppm:s like this if gas is saturated:
X = Pa / (Ptot - Pa) * 10 ^ 6
Where X is concentration in ppm, Pa is partial pressure of alcohol and Ptot is atmospheric pressure.
I'm just wondering if It should only be calculated from Ptot not Ptot-Pa.
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I've solved out my problem. One has to take into account, that gas concentration is not only a function of gas/air ratio (at given temperature and partial pressure), but also is a function of molecular weigt of investigated gases. So the final equation i've used was:
C [ppm] = (mgas/mair)*(Mair/Mgas)*106
where mgas and mair stands for weight of gas and air in chamber respectively, and Met and Mair stands for molar weight of gas and air respectively.